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0875-Koko-Eating-Bananas.md

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Koko Eating Bananas

Intuition

Koko the gorilla loves to eat bananas. Given N piles of bananas, where each pile i contains piles[i] bananas, Koko can eat bananas at a certain speed k bananas per hour. If a pile contains fewer bananas than her eating speed, she will finish that pile and wait for the next hour. The goal is to determine the minimum integer value of k such that Koko can eat all the bananas within h hours. This problem can be efficiently solved using a binary search approach due to the constraints and properties of the problem.

Approach

The key to solving this problem efficiently is to recognize that it can be framed as a binary search problem. We need to find the smallest possible value of k such that Koko can eat all the bananas in h hours. Here’s the approach step-by-step:

  1. Define Search Range:
    • The minimum eating speed k can be 1 (i.e., Koko eats at least one banana per hour).
    • The maximum eating speed k can be the size of the largest pile (i.e., Koko eats the entire largest pile in one hour).
  2. Binary Search:
    • Use binary search to find the minimum k that allows Koko to eat all the bananas within h hours.
    • Calculate the middle point in the current range.
    • Determine the total number of hours required for Koko to eat all bananas at the speed of the middle point.
  3. Ceiling Division: Use ceiling division to calculate the hours required to finish each pile since Koko can eat k bananas per hour but might not completely finish a pile in a single hour if pile % k > 0.
  4. Update Search Range:
    • If the total hours required is less than or equal to h, move the search range to the left (i.e., try a smaller eating speed).
    • If the total hours required is greater than h, move the search range to the right (i.e., try a larger eating speed).
  5. Terminate and Return: When the binary search completes, the left pointer will point to the minimum eating speed k.

Complexity

  • Time Complexity: O(Nlog(max(piles)))
    • Binary search runs in O(log(max(piles))).
    • Each step involves calculating the total hours, which takes O(N).
  • Space Complexity: O(1) The algorithm uses a constant amount of extra space.

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